JEE Main 2015 · 11 Apr · Shift 1 · Chemistry Q38

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At 298 K, the standard reduction potentials are 1.51 V for $\ce{MnO4- | Mn^2+}$ , 1.36 V for $\ce{Cl2 | Cl-}$ , 1.07 V for $\ce{Br2 | Br-}$ , and 0.54 V for $\ce{I2 | I-}$ . At pH = 3, permanganate is expected to oxidize : $\left(\dfrac{RT}F=0.059\text{ V}\right)$

\ce{Cl-}

\ce{Br-}

and

\ce{I-}

\ce{Cl-}

and

\ce{Br-}

\ce{Br-}

and

\ce{I-}

\ce{I-}

only

Answer and worked solution

Official answer: C

For $\ce{MnO4- + 8H+ + 5e- -> Mn^2+ + 4H2O}$ at pH 3:
$E=1.51-\frac{0.059}5\times8\times3\approx1.23\text{ V}.$
That exceeds 1.07 V (Br₂/Br⁻) and 0.54 V (I₂/I⁻) but not 1.36 V (Cl₂/Cl⁻). So permanganate oxidises $\ce{Br-}$ and $\ce{I-}$ .

Topics:Nernst equation

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JEE Main 2015 · 11 Apr · Shift 1, Chemistry question 38